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Hint: Here we solve the problem by finding the given set is a function or not.
Given $g = \{ (1,1),(2,3),(3,5),(4,7)\} $
Here, if we observe in the given set each element of domain has unique image so from this we can say the g is function.
Here the function g is described by a formula $g(x) = \alpha x + \beta $
Now let us substitute $x = 1$ in the formula $g(x) = \alpha x + \beta $
$\
g(1) = \alpha (1) + \beta \\
g(1) = \alpha + \beta \to 1 \\
\ $
We know that $g(1) = 1$ from the given function g
So we write the equation as
$\alpha + \beta = 1 \to 1$
Now again let us substitute $x = 2$in the formula $g(x) = \alpha x + \beta $
$\
g(2) = \alpha (2) + \beta \\
g(2) = 2\alpha + \beta \\
\ $
And we know that $g(2) = 3$from the given function g
So we can write the equation as
$2\alpha + \beta = 3$$ \to 2$
Now let us solve equation 1 and 2
We get $\alpha = 2$ and $\beta = 1$
Now we can write the formula as $g(x) = 2x - 1$
$\therefore g(x) = 2x - 1$
NOTE: We have used the domain values of g function to get the formula of g(x).
Given $g = \{ (1,1),(2,3),(3,5),(4,7)\} $
Here, if we observe in the given set each element of domain has unique image so from this we can say the g is function.
Here the function g is described by a formula $g(x) = \alpha x + \beta $
Now let us substitute $x = 1$ in the formula $g(x) = \alpha x + \beta $
$\
g(1) = \alpha (1) + \beta \\
g(1) = \alpha + \beta \to 1 \\
\ $
We know that $g(1) = 1$ from the given function g
So we write the equation as
$\alpha + \beta = 1 \to 1$
Now again let us substitute $x = 2$in the formula $g(x) = \alpha x + \beta $
$\
g(2) = \alpha (2) + \beta \\
g(2) = 2\alpha + \beta \\
\ $
And we know that $g(2) = 3$from the given function g
So we can write the equation as
$2\alpha + \beta = 3$$ \to 2$
Now let us solve equation 1 and 2
We get $\alpha = 2$ and $\beta = 1$
Now we can write the formula as $g(x) = 2x - 1$
$\therefore g(x) = 2x - 1$
NOTE: We have used the domain values of g function to get the formula of g(x).
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